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Chapter 6 Class 12 Application of Derivatives

Serial order wise

Last updated at April 19, 2021 by Teachoo

Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4𝑟/3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let ∠ BOC = θ Now, AC = AO + OC H = r + r cos θ H = r (1 + cos θ) And, R = r sin θ We need to maximize volume of cone. Volume of the cone is V = 1/3 𝜋𝑅^2 𝐻 V = 1/3 𝜋𝑟^2 sin^2"θ" r (1 + cos θ) V = 𝟏/𝟑 𝝅𝒓^𝟑 〖𝒔𝒊𝒏〗^𝟐"θ" (1 + cos θ) Finding 𝒅𝑽/𝒅𝜽 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 (𝑑[sin^2〖θ 〗 (1 + cosθ))/𝑑𝜃 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 [2 sin〖"θ" cos〖"θ" (1+cos〖"θ" )+sin^2〖"θ" (−sin〖"θ" )〗 〗 〗 〗 〗 ] 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 (2 sin〖"θ" cos〖"θ" (1+cos〖"θ" )−sin^3"θ" 〗 〗 〗 ) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 (2 sin〖"θ" cos〖"θ" (1+cos〖"θ" )−sin^3"θ" 〗 〗 〗 ) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (2 cos〖"θ" +〖2 cos^2〗〖"θ" −〖𝒔𝒊𝒏〗^𝟐"θ" 〗 〗 ) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (2 cos〖"θ" +〖2 cos^2〗〖"θ" −(𝟏−〖𝒄𝒐𝒔〗^𝟐"θ" 〗 〗)) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (𝟑 〖𝒄𝒐𝒔〗^𝟐〖"θ" +〖𝟐 𝒄𝒐𝒔〗〖"θ" −𝟏〗 〗 ) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (𝟑 〖𝒄𝒐𝒔〗^𝟐〖"θ" +〖𝟑 𝒄𝒐𝒔〗𝜽 〗 −𝒄𝒐𝒔𝜽 −𝟏) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 〖𝑐𝑜𝑠 〗"θ" (cos𝜃+1)−1(cos𝜃+1)) 𝑑𝑉/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 𝑐𝑜𝑠〖"θ" −1〗 )(𝑐𝑜𝑠〖"θ" +1〗 ) Putting 𝒅𝑽/𝒅𝜽 = 0 1/3 𝜋𝑟^3 sin θ (𝑐𝑜𝑠〖"θ" +1〗 ) (3 𝑐𝑜𝑠〖"θ" −1〗 ) = 0 𝐬𝐢𝐧 θ (cos θ + 1) (3 cos "θ" − 1) = 0 𝐬𝐢𝐧 θ = 0 θ = 0° θ cannot be 0° for cone 𝐜𝐨𝐬 θ + 1 = 0 cos θ = −1 For cone, 0° < θ < 90° & cos θ is negative in 2nd & 3rd quadrant. So cos θ = −1 is not possible 𝟑 𝐜𝐨𝐬 θ − 1 = 0 cos θ = 1/3 θ = cos−1 1/3 cos θ = 𝟏/𝟑 is possible So, 𝒄𝒐𝒔"θ" = 𝟏/𝟑 Thus, H = r (1 + cos θ) H = r ("1 + " 1/3) H = 𝟒𝒓/𝟑 Finding (𝒅^𝟐 𝑽)/(𝒅𝜽^𝟐 ) 𝑑𝑉/𝑑𝜃 = (1/3 𝜋𝑟^3 sin𝜃 (3 cos2 𝜃 + 2 cos𝜃− 1)) = 1/3 𝜋𝑟^3 [cos𝜃 (3 cos^2𝜃+2 cos𝜃−1)+sin𝜃 (6 cos𝜃 (−𝑠𝑖𝑛)−2 sin𝜃 )] = 1/3 𝜋𝑟^3 [3 cos^3𝜃+2 cos^2𝜃−cos𝜃+sin𝜃 (−6 cos𝜃 sin𝜃−2 sin𝜃 )] = 1/3 𝜋𝑟^3 [3 cos^2𝜃+2 cos^2𝜃−cos𝜃−6 sin^2𝜃 cos𝜃−2 sin^2𝜃 ] Now, 𝐜𝐨𝐬𝜽=𝟏/𝟑 And sin2 "θ" = 1 − cos2 "θ" = 1 − (1/3)^2= 1 − 1/9 = 𝟖/𝟗 Putting values in (𝒅^𝟐 𝑽)/(𝒅𝜽^𝟐 ) (𝑑^2 𝑉)/(𝑑𝜃^2 ) = 1/3 𝜋𝑟^3 [3(1/3)^3+2(1/3)^2−1/3−6(1/3)(8/9)−2(8/9)] = 1/3 𝜋𝑟^3 [3(1/27)+2(1/9)−1/3−6(8/27)−2(8/9)] = 1/3 𝜋𝑟^3 [1/9+ 2/9−1/3−16/9−16/9] = 1/3 𝜋𝑟^3 [(−32)/9] = (−𝟑𝟐𝝅𝒓^𝟑)/𝟐𝟕 Thus, (𝑑^2 𝑉)/(𝑑𝜃^2 ) < 0 for 𝐜𝐨𝐬𝜽=𝟏/𝟑 So, V is maximum for 𝐜𝐨𝐬𝜽=𝟏/𝟑 Hence, altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 𝟒𝒓/𝟑 .